Specific force ($\mathbf{f}=\mathbf{a}-\mathbf{g}$)

Specific force is the nongravitational force per unit mass experienced by a body. In a Newtonian local model, neglecting Earth's rotation, let $\mathbf{a}$ be its coordinate acceleration and $\mathbf{g}$ the local gravitational acceleration vector. Newton's second law givesSpecific force has units $\mathrm{m}{\mathrm{s}}^{-2}$. An ideal accelerometer senses this quantity along its own axes, rather than coordinate acceleration alone. The Android sensor documentation illustrates the distinction with a nonzero stationary-table reading and approximately zero reading during free fall.^[ref]

Choose an upward unit vector $\hat{\mathbf{z}}$ and $\mathbf{g}=-g\hat{\mathbf{z}}$. A stationary supported instrument has $\mathbf{a}=0$ and therefore $\mathbf{f}=+g\hat{\mathbf{z}}$: support supplies the upward nongravitational force. In ideal free fall, $\mathbf{a}=\mathbf{g}$ and $\mathbf{f}=0$. Gravity has not disappeared; the instrument and its proof mass accelerate together.

For an original elevator example with $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, an upward coordinate acceleration of $2.00\mathrm{m}{\mathrm{s}}^{-2}$ gives upward specific force $11.81\mathrm{m}{\mathrm{s}}^{-2}$. Moving upward at constant velocity instead gives $9.81\mathrm{m}{\mathrm{s}}^{-2}$. The reading depends on acceleration and support, not the direction of velocity.

If a rotation matrix $R$ maps instrument-axis components to world-axis components, an ideal body-axis readout is $R^{\mathsf{T}}(\mathbf{a}-\mathbf{g})$. Reconstructing $\mathbf{a}$ therefore also requires orientation and a gravity model. A sustained horizontal acceleration can resemble a tilted gravity-related reading; an accelerometer alone does not generally separate those causes. In this local slow-motion setting the sensed support acceleration connects to proper acceleration.